Definition: Manifold

A (topological) **manifold** of dimension $m$ is a topological space $M$ s.t. each point $P\in M$ admits a neighborhood $U \subset M$ homeomorphic to an open subset $V \subseteq \mathbb{R}^m$

i.e. $M$ is locally $m$-dimensionally euclidian.

$U$, together with a homeomorphism $\phi: U \to V \subseteq \mathbb{R}^m$ is called a chart. A collection of charts covering $M$ is called an atlas.

Fixing a chart $U$ is equivalent to defining local coordinates $(x_1, \dots, x_m)$ on $U$, with $\phi(P) = (x_1, \dots, x_m) \in V \subseteq \mathbb{R}^m$.

We also assume the following conditions:

- $M$ is hausdorff, i.e. you can distinguish any two points, i.e. any two distinct points have disjoint neighborhoods.

- $M$ admits a countable atlas

Smooth: of class $C^{\infty}$ (i.e. infinitely differentiable)

We want smooth manifolds because we want to work with local coordinates in our manifold like we may in $\mathbb{R}^m$. The problem is if some points belong to several charts, we have several choices for local coordinates and we want them to match/correspond.

A manifold is smooth if the transition functions between any two local coordinate systems are smooth.

That is, given two coordinate maps $\phi: V \to \mathbb{R}^m$ and $\phi': V' \to \mathbb{R}^m$, the transition function $\phi'\circ\phi^{-1}$, which takes points in $V$ to points in $V'$, should be smooth.

If $M$ is smooth, then the transition functions should be smooth.

Examples of manifolds:

- Vector space $\mathbb{R}^n$

- Any open set of $\mathbb{R}^n$

- Graph of a smooth function (map)

- Two dimensional surfaces

- Spheres $S^n$ and projective spaces $\mathbb{RP}^n$

- Lie Groups (of course)

- Homogeneous spaces of Lie groups

- Subsets in $\mathbb{R}^n$ given by a system of equations (with certain "regularity" conditions, guaranteed by implicit function theorem)

Theorem: Implicit Function Theorem

In $\mathbb{R}^n$, consider a system of equations $f_1 = 0, \dots, f_k = 0$ where $f_i$ are smooth and $k <= n$. Let $M \subseteq \mathbb{R}^n$ be the set of solutions.

Then, the rank of the jacobi matrix is maximal (that is, $=k$) at any point $P \in M$.

In other words, because we can think of each row of the matrix as a vector, and because we know that it must be linearly independent, we know that each system of equations must be linearly independent

Theorem: If the regularity condition holds, then $M$ carries the natural structure of a smooth manifold of dimension $n - k$.

Example:

Consider the group $O(3)$ as a subset in $\mathbb{R}^9$ (3x3 matrices). The condition $\mathbf{AA}^T = \text{id}$ is a matrix equation which is equivalent to a system of 6 usual equations:

$\begin{align*} &f_1: a_{11}^2 + a_{12}^2 + a_{13}^2 = 1 \\ &f_2: a_{11}a_{21} + a_{12}a_{22} + a_{13}a_{23} = 0 \\ &f_3: a_{11}a_{31} + a_{12}a_{32} + a_{13}a_{33} = 0 \\ &f_4: a_{21}^2 + a_{22}^2 + a_{23}^2 = 1 \\ &f_5: a_{21}a_{31} + a_{22}a_{32} + a_{23}a_{33} = 0 \\ &f_6: a_{31}^2 + a_{32}^2 + a_{33}^2 = 1 \\ \end{align*}$

Then, if we find the jacobi matrix, we see that it's a block-triangular matrix. We see that if we add the ranks of each block, we get 6, so the matrix has rank 6. Therefore, $O(3)$ is a $9 - 6 = 3$ dimensional smooth manifold.

Another example:

Consider $V$, a cone, in $\mathbb{R}^3$ given by $x^2 + y^2 - z^2 = 0$.

At most points, the regularity condition $df \neq 0$ holds. But at $(0, 0, 0)$, the derivative is $0$ so the condition fails and $V$, as a whole, is not a smooth manifold.

Looking at $V$ itself, it's easy to see that $V$ isn't smooth because $(0, 0, 0)$ doesn't have a neighborhood homeomorphic to a $2$-disk.

A continuous map $F: M \to N$ between smooth manifolds is smooth if it is so in local coordinates.

More precisely, if $P \in M$ and $Q = F(P) \in N$ its image, with coordinates $(x_1, \dots, x_m)$ and $(y_1, \dots, y_n)$ in neighborhoods of $P$ and $Q$ resp, then $F$ can be written using these local coordinates:

$(y_1, \dots, y_n) = F(x_1, \dots, x_m)$, i.e. $F = (f_1(x_1, \dots, x_m), f_2(x_1, \dots, x_m), \dots, f_n(x_1, \dots, x_m))$

Smoothness of $F$ means that all $f_1, \dots, f_n$ are smooth.

Since the transition functions between charts are smooth, the above definition does not depend on the choice of local coordinates.

Definition: Diffeomorphism

A **diffeomorphism** $F: M \to N$ is a smooth bijective map such that its inverse is also smooth. #-> So basically an isomorphism in the context of manifolds

In the simple sense, if we have a manifold $M$ embedded in $\mathbb{R}^n$, a tangent vector at a point $P \in M$ is just a tangent vector to a smooth curve passing through $P$.

The set of all vectors at $P$ is the tangent space to $M$ at $P$.

If we let $\gamma(t)$ be a smooth curve in $M$, i.e. a smooth map from $(-\epsilon, \epsilon) \to M$, then in local coordinates $\gamma(t) = (x_1(t), \dots, x_m(t))$.

Then, the tangent vector to gamma at point $P = \gamma(0)$ is defined to be simply $\frac{d}{dt}\gamma(0) = \left(\frac{d}{dt}x_1(0), \frac{d}{dt}x_2(0), \dots, \frac{d}{dt}x_m(0)\right)$

So, in local coordinates, any tangent vector is given as an $m$-tuple.

The problem is when we change local coordinate systems, as our function will be defined by new functions. Then, the new tangent vector in the new coordinates becomes

$\frac{d}{dt}\gamma' = \left(\frac{d}{dt}x_1', \dots, \frac{d}{dt}x_m'\right) = \left(\sum_{i=1}^m \frac{\partial}{\partial x_i} x_1' \cdot \frac{d}{dt}x_i, \dots, \sum_{i=1}^m \frac{\partial}{\partial x_i} x_m' \cdot \frac{d}{dt}x_i\right)$

Then, if the original tangent vector is $(\xi_1, \xi_2, \dots, \xi_m)$, then the same vector in the new coordinate system becomes

$(\xi_1', \dots, \xi_m') = \left(\sum_{i=1}^m \frac{\partial}{\partial x_i} x_1'\cdot \xi_i, \dots, \sum_{i=1}^m \frac{\partial}{\partial x_i} x_m'\cdot \xi_i\right)$

Definition: Tangent Vector

A **tangent vector** $\xi$ at point $P$ is defined in any local coordinate system as an $m$-tuple.

In addition, it is required that the transformation law for the components of $\xi$ is given by the above equation.